IODS course project
Margot de Bazillac
Introduction to Open Data Science - Course Project
About the project
date()## [1] "Mon Nov 27 17:22:45 2023"Heads-up: I am new with R. During this first week, I
studied the 4 first chapters of the book made on R studio and watched
the given tutorial videos.
I have learned how to:
- set my directory
- use the R project mode, the R markdown file and github commit
- install and use libraries
- identify and use the different data types including the factors
- use the pipe %>% (it means “and then…”)
- open data set, read data set, read the structure of the data set
- select rows, columns and match text or rename column name
- filter data based on variable values
- create a table / tibble
- create an object that is reusable later
- use the time format
- add columns with text, int, calculations, categories etc.
- group data with one or many variables to then perform
calculation
- combine all data types to create a string with the function
paste()
- check the NA and remove the NA to do calculations
- summarize to sum up data per variable value
- manipulate table from wide to long and long to wide
- combine 2 tables (functions: bind, inner_join, full_join, left_join,
right_join)
- use the most common arithmetic functions and use percentage
- rank and reorder value
- create plots (bar, col, line, point, box, histogram)
- play with the different style and aesthetic possibilities
- add labels to plots
- combine plots together
With this tool I discovered how easy and fast it can be to manipulate
and visualize data from a big data set.
Conclusion of Week 1:
The book is really well made. The style is easy to follow and
understand. We can learn at our own pace and try the examples in R
studio to assimilate the concepts well.
I am new with R and I find it hard to learn so many new different
concepts in a week. I would have liked to have more exercises, without
the solutions directly written under and with a tool to correct our
lines of code. Finally, I understand it is not necessary to know by
heart every function, but I would like to understand well the
capabilities of this tool by practicing and doing more real life
exercises.
Lines of code to summarize learning in week 1:
# Get the data set
library(tidyverse)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.3 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.0
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorslibrary(scales)##
## Attaching package: 'scales'
##
## The following object is masked from 'package:purrr':
##
## discard
##
## The following object is masked from 'package:readr':
##
## col_factorgbd_full <- read_csv("https://raw.githubusercontent.com/KimmoVehkalahti/RHDS/master/data/global_burden_disease_cause-year-sex-income.csv")## Rows: 168 Columns: 5
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (3): cause, sex, income
## dbl (2): year, deaths_millions
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.# This is a code I wrote to know the pourcentage of deaths per cause for each year.
exercise4 <- gbd_full %>%
group_by(year, cause) %>%
summarise(total_per_cause= sum(deaths_millions)) %>%
group_by(year) %>%
mutate(total_per_year =sum(total_per_cause)) %>%
mutate(percentage = percent(total_per_cause/total_per_year)) %>%
select(year,cause,percentage) %>%
pivot_wider(names_from = cause, values_from = percentage)## `summarise()` has grouped output by 'year'. You can override using the
## `.groups` argument.exercise4## # A tibble: 7 × 4
## # Groups: year [7]
## year `Communicable diseases` Injuries `Non-communicable diseases`
## <dbl> <chr> <chr> <chr>
## 1 1990 33% 9% 58%
## 2 1995 31% 9% 60%
## 3 2000 29% 9% 62%
## 4 2005 27% 9% 64%
## 5 2010 24% 9% 67%
## 6 2015 20% 8% 72%
## 7 2017 19% 8% 73%# This is a code I wrote to know the number of deaths per sex and income and see the ratio Male / Female for each income type.
gbd_full %>%
filter(year ==1990) %>%
group_by(sex,income) %>%
summarize(deaths_per_sex_income = sum(deaths_millions)) %>%
pivot_wider(names_from = sex, values_from = deaths_per_sex_income) %>%
mutate(diff_M_F = Male/Female)## `summarise()` has grouped output by 'sex'. You can override using the `.groups`
## argument.## # A tibble: 4 × 4
## income Female Male diff_M_F
## <chr> <dbl> <dbl> <dbl>
## 1 High 4.14 4.46 1.08
## 2 Low 2.22 2.57 1.16
## 3 Lower-Middle 8.47 9.83 1.16
## 4 Upper-Middle 6.68 7.95 1.19Chapter 2: Regression and model validation
date()## [1] "Mon Nov 27 17:22:48 2023"# set directory
setwd("/Users/margot/Desktop/Desktop - MacBook Pro de MARGOT/Open data with R 2023/IODS-project")Thoughts about this week 2:
After reading all the chapters 1-7, I am now more confident to use R studio. I also understand the language better and I can do research on the web to use new function that I did not know.
It is very exciting to see how efficient is this tool and to think about all the analyzes we can do. I am an open university student and I can already see how to use this tool at work :).
Exercise: WRANGLING Please find the script file in the Github: create_learning2014_week2
Exercise: DATA ANALYSIS
QUESTION 1:
# Using the table from the course.
csv_table_read <- read.table("https://raw.githubusercontent.com/KimmoVehkalahti/Helsinki-Open-Data-Science/master/datasets/learning2014.txt", sep = ",", header = T)
# library
library("tidyverse")
library("finalfit")
library("broom")
csv_table_read## gender age attitude deep stra surf points
## 1 F 53 3.7 3.583333 3.375 2.583333 25
## 2 M 55 3.1 2.916667 2.750 3.166667 12
## 3 F 49 2.5 3.500000 3.625 2.250000 24
## 4 M 53 3.5 3.500000 3.125 2.250000 10
## 5 M 49 3.7 3.666667 3.625 2.833333 22
## 6 F 38 3.8 4.750000 3.625 2.416667 21
## 7 M 50 3.5 3.833333 2.250 1.916667 21
## 8 F 37 2.9 3.250000 4.000 2.833333 31
## 9 M 37 3.8 4.333333 4.250 2.166667 24
## 10 F 42 2.1 4.000000 3.500 3.000000 26
## 11 M 37 3.9 3.583333 3.625 2.666667 31
## 12 F 34 3.8 3.833333 4.750 2.416667 31
## 13 F 34 2.4 4.250000 3.625 2.250000 23
## 14 F 34 3.0 3.333333 3.500 2.750000 25
## 15 M 35 2.6 4.166667 1.750 2.333333 21
## 16 F 33 4.1 3.666667 3.875 2.333333 31
## 17 F 32 2.6 4.083333 1.375 2.916667 20
## 18 F 44 2.6 3.500000 3.250 2.500000 22
## 19 M 29 1.7 4.083333 3.000 3.750000 9
## 20 F 30 2.7 4.000000 3.750 2.750000 24
## 21 M 27 3.9 3.916667 2.625 2.333333 28
## 22 M 29 3.4 4.000000 2.375 2.416667 30
## 23 F 31 2.7 4.000000 3.625 3.000000 24
## 24 F 37 2.3 3.666667 2.750 2.416667 9
## 25 F 26 3.7 3.666667 1.750 2.833333 26
## 26 F 26 4.4 4.416667 3.250 3.166667 32
## 27 M 30 4.1 3.916667 4.000 3.000000 32
## 28 F 33 3.7 3.750000 3.625 2.000000 33
## 29 F 33 2.5 3.250000 2.875 3.500000 29
## 30 M 28 3.0 3.583333 3.000 3.750000 30
## 31 M 26 3.4 4.916667 1.625 2.500000 19
## 32 F 27 3.2 3.583333 3.250 2.083333 23
## 33 F 25 2.0 2.916667 3.500 2.416667 19
## 34 F 31 2.4 3.666667 3.000 2.583333 12
## 35 M 20 4.2 4.500000 3.250 1.583333 10
## 36 F 39 1.6 4.083333 1.875 2.833333 11
## 37 M 38 3.1 3.833333 4.375 1.833333 20
## 38 M 24 3.8 3.250000 3.625 2.416667 26
## 39 M 26 3.8 2.333333 2.500 3.250000 31
## 40 M 25 3.3 3.333333 1.250 3.416667 20
## 41 F 30 1.7 4.083333 4.000 3.416667 23
## 42 F 25 2.5 2.916667 3.000 3.166667 12
## 43 M 30 3.2 3.333333 2.500 3.500000 24
## 44 F 48 3.5 3.833333 4.875 2.666667 17
## 45 F 24 3.2 3.666667 5.000 2.416667 29
## 46 F 40 4.2 4.666667 4.375 3.583333 23
## 47 M 25 3.1 3.750000 3.250 2.083333 28
## 48 F 23 3.9 3.416667 4.000 3.750000 31
## 49 F 25 1.9 4.166667 3.125 2.916667 23
## 50 F 23 2.1 2.916667 2.500 2.916667 25
## 51 M 27 2.5 4.166667 3.125 2.416667 18
## 52 M 25 3.2 3.583333 3.250 3.000000 19
## 53 M 23 3.2 2.833333 2.125 3.416667 22
## 54 F 23 2.6 4.000000 2.750 2.916667 25
## 55 F 23 2.3 2.916667 2.375 3.250000 21
## 56 F 45 3.8 3.000000 3.125 3.250000 9
## 57 F 22 2.8 4.083333 4.000 2.333333 28
## 58 F 23 3.3 2.916667 4.000 3.250000 25
## 59 M 21 4.8 3.500000 2.250 2.500000 29
## 60 M 21 4.0 4.333333 3.250 1.750000 33
## 61 F 21 4.0 4.250000 3.625 2.250000 33
## 62 F 21 4.7 3.416667 3.625 2.083333 25
## 63 F 26 2.3 3.083333 2.500 2.833333 18
## 64 F 25 3.1 4.583333 1.875 2.833333 22
## 65 F 26 2.7 3.416667 2.000 2.416667 17
## 66 M 21 4.1 3.416667 1.875 2.250000 25
## 67 F 23 3.4 3.416667 4.000 2.833333 28
## 68 F 22 2.5 3.583333 2.875 2.250000 22
## 69 F 22 2.1 1.583333 3.875 1.833333 26
## 70 F 22 1.4 3.333333 2.500 2.916667 11
## 71 F 23 1.9 4.333333 2.750 2.916667 29
## 72 M 22 3.7 4.416667 4.500 2.083333 22
## 73 M 23 3.2 4.833333 3.375 2.333333 21
## 74 M 24 2.8 3.083333 2.625 2.416667 28
## 75 F 22 4.1 3.000000 4.125 2.750000 33
## 76 F 23 2.5 4.083333 2.625 3.250000 16
## 77 M 22 2.8 4.083333 2.250 1.750000 31
## 78 M 20 3.8 3.750000 2.750 2.583333 22
## 79 M 22 3.1 3.083333 3.000 3.333333 31
## 80 M 21 3.5 4.750000 1.625 2.833333 23
## 81 F 22 3.6 4.250000 1.875 2.500000 26
## 82 F 23 2.6 4.166667 3.375 2.416667 12
## 83 M 21 4.4 4.416667 3.750 2.416667 26
## 84 M 22 4.5 3.833333 2.125 2.583333 31
## 85 M 29 3.2 3.333333 2.375 3.000000 19
## 86 F 29 3.9 3.166667 2.750 2.000000 30
## 87 F 21 2.5 3.166667 3.125 3.416667 12
## 88 M 28 3.3 3.833333 3.500 2.833333 17
## 89 F 21 3.3 4.250000 2.625 2.250000 18
## 90 F 30 3.0 3.833333 3.375 2.750000 19
## 91 F 21 2.9 3.666667 2.250 3.916667 21
## 92 M 23 3.3 3.833333 3.000 2.333333 24
## 93 F 21 3.3 3.833333 4.000 2.750000 28
## 94 F 21 3.5 3.833333 3.500 2.750000 17
## 95 F 20 3.6 3.666667 2.625 2.916667 18
## 96 M 22 3.7 4.333333 2.500 2.083333 17
## 97 M 21 4.2 3.750000 3.750 3.666667 23
## 98 M 21 3.2 4.166667 3.625 2.833333 26
## 99 F 20 5.0 4.000000 4.125 3.416667 28
## 100 M 22 4.7 4.000000 4.375 1.583333 31
## 101 F 20 3.6 4.583333 2.625 2.916667 27
## 102 F 20 3.6 3.666667 4.000 3.000000 25
## 103 M 24 2.9 3.666667 2.750 2.916667 23
## 104 F 20 3.5 3.833333 2.750 2.666667 21
## 105 F 19 4.0 2.583333 1.375 3.000000 27
## 106 F 21 3.5 3.500000 2.250 2.750000 28
## 107 F 21 3.2 3.083333 3.625 3.083333 23
## 108 F 22 2.6 4.250000 3.750 2.500000 21
## 109 F 25 2.0 3.166667 4.000 2.333333 25
## 110 F 21 2.7 3.083333 3.125 3.000000 11
## 111 F 22 3.2 4.166667 3.250 3.000000 19
## 112 F 25 3.3 2.250000 2.125 4.000000 24
## 113 F 20 3.9 3.333333 2.875 3.250000 28
## 114 M 24 3.3 3.083333 1.500 3.500000 21
## 115 F 20 3.0 2.750000 2.500 3.500000 24
## 116 M 21 3.7 3.250000 3.250 3.833333 24
## 117 F 20 2.5 4.000000 3.625 2.916667 20
## 118 F 20 2.9 3.583333 3.875 2.166667 19
## 119 M 31 3.9 4.083333 3.875 1.666667 30
## 120 F 20 3.6 4.250000 2.375 2.083333 22
## 121 F 22 2.9 3.416667 3.000 2.833333 16
## 122 F 22 2.1 3.083333 3.375 3.416667 16
## 123 M 21 3.1 3.500000 2.750 3.333333 19
## 124 M 22 4.0 3.666667 4.500 2.583333 30
## 125 F 21 3.1 4.250000 2.625 2.833333 23
## 126 F 21 2.3 4.250000 2.750 3.333333 19
## 127 F 21 2.8 3.833333 3.250 3.000000 18
## 128 F 21 3.7 4.416667 4.125 2.583333 28
## 129 F 20 2.6 3.500000 3.375 2.416667 21
## 130 F 21 2.4 3.583333 2.750 3.583333 19
## 131 F 25 3.0 3.666667 4.125 2.083333 27
## 132 M 21 2.8 2.083333 3.250 4.333333 24
## 133 F 24 2.9 4.250000 2.875 2.666667 21
## 134 F 20 2.4 3.583333 2.875 3.000000 20
## 135 M 21 3.1 4.000000 2.375 2.666667 28
## 136 F 20 1.9 3.333333 3.875 2.166667 12
## 137 F 20 2.0 3.500000 2.125 2.666667 21
## 138 F 18 3.8 3.166667 4.000 2.250000 28
## 139 F 21 3.4 3.583333 3.250 2.666667 31
## 140 F 19 3.7 3.416667 2.625 3.333333 18
## 141 F 21 2.9 4.250000 2.750 3.500000 25
## 142 F 20 2.3 3.250000 4.000 2.750000 19
## 143 M 21 4.1 4.416667 3.000 2.000000 21
## 144 F 20 2.7 3.250000 3.375 2.833333 16
## 145 F 21 3.5 3.916667 3.875 3.500000 7
## 146 F 20 3.4 3.583333 3.250 2.500000 21
## 147 F 18 3.2 4.500000 3.375 3.166667 17
## 148 M 22 3.3 3.583333 4.125 3.083333 22
## 149 F 22 3.3 3.666667 3.500 2.916667 18
## 150 M 24 3.5 2.583333 2.000 3.166667 25
## 151 F 19 3.2 4.166667 3.625 2.500000 24
## 152 F 20 3.1 3.250000 3.375 3.833333 23
## 153 F 20 2.8 4.333333 2.125 2.250000 23
## 154 F 17 1.7 3.916667 4.625 3.416667 26
## 155 M 19 1.9 2.666667 2.500 3.750000 12
## 156 F 20 3.5 3.083333 2.875 3.000000 32
## 157 F 20 2.4 3.750000 2.750 2.583333 22
## 158 F 20 2.1 4.166667 4.000 3.333333 20
## 159 F 20 2.9 4.166667 2.375 2.833333 21
## 160 F 19 1.9 3.250000 3.875 3.000000 23
## 161 F 19 2.0 4.083333 3.375 2.833333 20
## 162 F 22 4.2 2.916667 1.750 3.166667 28
## 163 M 35 4.1 3.833333 3.000 2.750000 31
## 164 F 18 3.7 3.166667 2.625 3.416667 18
## 165 F 19 3.6 3.416667 2.625 3.000000 30
## 166 M 21 1.8 4.083333 3.375 2.666667 19# analyze the structure of the dataset
str(csv_table_read)## 'data.frame': 166 obs. of 7 variables:
## $ gender : chr "F" "M" "F" "M" ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...# analyze the dimension of the dataset
dim(csv_table_read)## [1] 166 7# Missing data? No data is missing.
ff_glimpse(csv_table_read)## $Continuous
## label var_type n missing_n missing_percent mean sd min
## age age <int> 166 0 0.0 25.5 7.8 17.0
## attitude attitude <dbl> 166 0 0.0 3.1 0.7 1.4
## deep deep <dbl> 166 0 0.0 3.7 0.6 1.6
## stra stra <dbl> 166 0 0.0 3.1 0.8 1.2
## surf surf <dbl> 166 0 0.0 2.8 0.5 1.6
## points points <int> 166 0 0.0 22.7 5.9 7.0
## quartile_25 median quartile_75 max
## age 21.0 22.0 27.0 55.0
## attitude 2.6 3.2 3.7 5.0
## deep 3.3 3.7 4.1 4.9
## stra 2.6 3.2 3.6 5.0
## surf 2.4 2.8 3.2 4.3
## points 19.0 23.0 27.8 33.0
##
## $Categorical
## label var_type n missing_n missing_percent levels_n levels
## gender gender <chr> 166 0 0.0 2 -
## levels_count levels_percent
## gender - -# summary statistics for each variable
missing_glimpse(csv_table_read)## label var_type n missing_n missing_percent
## gender gender <chr> 166 0 0.0
## age age <int> 166 0 0.0
## attitude attitude <dbl> 166 0 0.0
## deep deep <dbl> 166 0 0.0
## stra stra <dbl> 166 0 0.0
## surf surf <dbl> 166 0 0.0
## points points <int> 166 0 0.0# Count per gender and percentage male / female
library("scales")
csv_table_read %>%
count(gender) %>%
mutate(total_percentage = n / nrow(csv_table_read)) %>%
mutate(total_percentage2 = percent(total_percentage))## gender n total_percentage total_percentage2
## 1 F 110 0.6626506 66%
## 2 M 56 0.3373494 34%# Mean and median for exercises points, and learning method per gender
summary(csv_table_read)## gender age attitude deep
## Length:166 Min. :17.00 Min. :1.400 Min. :1.583
## Class :character 1st Qu.:21.00 1st Qu.:2.600 1st Qu.:3.333
## Mode :character Median :22.00 Median :3.200 Median :3.667
## Mean :25.51 Mean :3.143 Mean :3.680
## 3rd Qu.:27.00 3rd Qu.:3.700 3rd Qu.:4.083
## Max. :55.00 Max. :5.000 Max. :4.917
## stra surf points
## Min. :1.250 Min. :1.583 Min. : 7.00
## 1st Qu.:2.625 1st Qu.:2.417 1st Qu.:19.00
## Median :3.188 Median :2.833 Median :23.00
## Mean :3.121 Mean :2.787 Mean :22.72
## 3rd Qu.:3.625 3rd Qu.:3.167 3rd Qu.:27.75
## Max. :5.000 Max. :4.333 Max. :33.00# The age varies from 17 to 55, mean is 25 and median 22. it suggests that there are some relatively higher values in the dataset
# The attitude varies from 1.4 to 5
# The points are from 7 to 33 and the mean is 22 and the median is 23. It suggests that there are some relatively lower values in the dataset
# we analyze the variables for both genders and females
# draw a scatter plot matrix of the variables in learning2014.
# [-1] excludes the first column (gender)
pairs(csv_table_read[-1])
# access the GGally and ggplot2 libraries
library(GGally)## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2library(ggplot2)
# create a more advanced plot matrix with ggpairs()
p <- ggpairs(csv_table_read, mapping = aes(), lower = list(combo = wrap("facethist", bins = 20)))
p
# some data shows that there could be correlation between some variables
# Relationship between points and attitudes
csv_table_read %>%
ggplot(aes(x= attitude, y= points)) +
geom_point() +
facet_wrap(~ gender) +
geom_smooth(method = "lm")## `geom_smooth()` using formula = 'y ~ x'
QUESTION 2, 3 and 4:
Female model
#female model
female_data <- csv_table_read %>%
filter(gender == "F")
View(female_data)
# Fit a multiple linear model for females. Let's check how points are influenced by age, attitude and deep learning approach
female_fitmodel <- lm(points ~ age + attitude + deep, data = female_data)
# In this model I want to check if age, attitude and deep impact points without impacting each other.
# summary of std, p value and
summary(female_fitmodel) ##
## Call:
## lm(formula = points ~ age + attitude + deep, data = female_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.058 -3.263 0.622 4.003 10.533
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 13.59029 4.28982 3.168 0.00201 **
## age -0.01743 0.06983 -0.250 0.80338
## attitude 3.40151 0.70837 4.802 5.19e-06 ***
## deep -0.27355 0.96270 -0.284 0.77685
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.353 on 106 degrees of freedom
## Multiple R-squared: 0.1791, Adjusted R-squared: 0.1559
## F-statistic: 7.71 on 3 and 106 DF, p-value: 0.0001043summary(female_fitmodel)$r.squared## [1] 0.1791183#"age," "attitude," and "deep" explains about 18% of the variation of "points"
# p value intercept: it is significant as very small (0.002) and seems to play a significant role in the regression model
# baseline of model in 13.59 (estimate), when no factors are taken into account.
# age is not significant and is not correlated with points
# deep is not significant and is not correlated with points
# attitude is significant and it seems to play a significant role on the points.
# for one point increase in the attitude, the points increase by 3.63 (estimate)
# High std shows that the estimate is not so precise. It could due to sample size.
# I decide to drop the deep and the age variables and keep only the attitude.
female_fitmodel2 <- lm(points ~ attitude, data = female_data)
summary(female_fitmodel2) ##
## Call:
## lm(formula = points ~ attitude, data = female_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.0557 -3.3486 0.6137 3.9819 10.3668
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 12.194 2.156 5.655 1.29e-07 ***
## attitude 3.389 0.701 4.835 4.44e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.307 on 108 degrees of freedom
## Multiple R-squared: 0.1779, Adjusted R-squared: 0.1703
## F-statistic: 23.38 on 1 and 108 DF, p-value: 4.442e-06tidy(female_fitmodel2)## # A tibble: 2 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 12.2 2.16 5.66 0.000000129
## 2 attitude 3.39 0.701 4.83 0.00000444summary(female_fitmodel2)$r.squared## [1] 0.1779266# p value is very low, same for the std, so this model is correct and justify the positive relation vs a positive attitude -> more points.
# rsquare is still quite low..
# The model doesn't provide a good fit for the data, and a significant portion of the variance is not explained. Is could be due to the sample size.
# autoplot: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
# Identify issues with my regression model, such as non-linearity, non-normality, or influential data points
# autoplot doesnt knit.
#autoplot(female_fitmodel)
#autoplot(female_fitmodel2)
# I use plot function and which to get the desired plots.
plot(female_fitmodel,which = c(1,2,5))
plot(female_fitmodel2,which = c(1,2,5))
# we observe non normality at the end and beginning of the line in qq plot
# both models show that there are some points that are high leverage indicated on the residuals vs leverageMale model
# male model
male_data <- csv_table_read %>%
filter(gender == "M")
View(male_data)
summary(male_data)## gender age attitude deep
## Length:56 Min. :19.0 Min. :1.700 Min. :2.083
## Class :character 1st Qu.:21.0 1st Qu.:3.100 1st Qu.:3.396
## Mode :character Median :24.0 Median :3.400 Median :3.792
## Mean :26.8 Mean :3.443 Mean :3.725
## 3rd Qu.:29.0 3rd Qu.:3.900 3rd Qu.:4.083
## Max. :55.0 Max. :4.800 Max. :4.917
## stra surf points
## Min. :1.250 Min. :1.583 Min. : 9.00
## 1st Qu.:2.375 1st Qu.:2.312 1st Qu.:20.00
## Median :3.000 Median :2.625 Median :23.50
## Mean :2.964 Mean :2.704 Mean :23.48
## 3rd Qu.:3.531 3rd Qu.:3.167 3rd Qu.:28.25
## Max. :4.500 Max. :4.333 Max. :33.00# Fit a multiple linear model for males. Let's check how points are influenced by age, attitude and deep learning approach
male_fitmodel <- lm(points ~ age + attitude + deep, data = male_data)
# summary of std, p value and
summary(male_fitmodel) ##
## Call:
## lm(formula = points ~ age + attitude + deep, data = male_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.8084 -3.3162 -0.0696 3.2195 9.9927
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 18.21897 6.06857 3.002 0.004112 **
## age -0.16602 0.08456 -1.963 0.054974 .
## attitude 4.31829 1.11699 3.866 0.000309 ***
## deep -1.38378 1.22006 -1.134 0.261916
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.271 on 52 degrees of freedom
## Multiple R-squared: 0.2718, Adjusted R-squared: 0.2298
## F-statistic: 6.47 on 3 and 52 DF, p-value: 0.000835tidy(male_fitmodel)## # A tibble: 4 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 18.2 6.07 3.00 0.00411
## 2 age -0.166 0.0846 -1.96 0.0550
## 3 attitude 4.32 1.12 3.87 0.000309
## 4 deep -1.38 1.22 -1.13 0.262summary(male_fitmodel)$r.squared ## [1] 0.2718164# similar results than for the female.
# All variables have a smaller p value than for in the female model.
# rsquare is higher as it explains 27% but it is still quite low. It could be due to the sample size.
# I decide to drop the deep and the age as variables and keep only the attitude.
male_fitmodel2 <- lm(points ~ attitude, data = male_data)
summary(male_fitmodel2) ##
## Call:
## lm(formula = points ~ attitude, data = male_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.6535 -2.9073 -0.5121 3.6974 10.2106
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 9.061 3.953 2.292 0.02581 *
## attitude 4.189 1.129 3.711 0.00049 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.411 on 54 degrees of freedom
## Multiple R-squared: 0.2032, Adjusted R-squared: 0.1884
## F-statistic: 13.77 on 1 and 54 DF, p-value: 0.0004897tidy(male_fitmodel2)## # A tibble: 2 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 9.06 3.95 2.29 0.0258
## 2 attitude 4.19 1.13 3.71 0.000490summary(male_fitmodel)$r.squared ## [1] 0.2718164# p value is very low, same for the std, so this model is correct and justify the positive relation vs a positive attitude -> more points.
# rsquare is higher as it explains 27% but it is still quite low
# The model doesn't provide a good fit for the data, and a significant portion of the variance is not explained. Is could be due to the sample size.
# autoplot: Residuals vs Fitted values, Normal QQ-plot and Residuals vs Leverage
# Identify issues with my regression model, such as non-linearity, non-normality, or influential data points
# autoplot doesnt knit !!
# autoplot(male_fitmodel)
# autoplot(male_fitmodel2)
# plot with the plot function
plot(male_fitmodel,which = c(1,2,5))
plot(male_fitmodel2,which = c(1,2,5))
#The red line in residuals vs fitted stays quite close to the 0 line which is good
# both models show non normality. it is observed at the beginning of the qq plot
# both models show that there are some points that are high leverage indicated on the residuals vs leverageother models tested:
test_fit1 <- csv_table_read %>%
lm(points ~ deep, data = .)
library(ggfortify)
summary(test_fit1)##
## Call:
## lm(formula = points ~ deep, data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.6913 -3.6935 0.2862 4.9957 10.3537
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 23.1141 3.0908 7.478 4.31e-12 ***
## deep -0.1080 0.8306 -0.130 0.897
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.913 on 164 degrees of freedom
## Multiple R-squared: 0.000103, Adjusted R-squared: -0.005994
## F-statistic: 0.01689 on 1 and 164 DF, p-value: 0.8967tidy(test_fit1) # p value is small and significant## # A tibble: 2 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 23.1 3.09 7.48 4.31e-12
## 2 deep -0.108 0.831 -0.130 8.97e- 1summary(test_fit1)$r.squared # too low## [1] 0.0001029919test_fit2 <- csv_table_read %>%
lm(points ~ deep * gender, data = .)
library(ggfortify)
summary(test_fit2)##
## Call:
## lm(formula = points ~ deep * gender, data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.3247 -3.3338 0.3369 4.6242 10.6787
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 22.36387 3.91828 5.708 5.33e-08 ***
## deep -0.01001 1.06032 -0.009 0.992
## genderM 2.67476 6.41719 0.417 0.677
## deep:genderM -0.40787 1.71487 -0.238 0.812
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.922 on 162 degrees of freedom
## Multiple R-squared: 0.00922, Adjusted R-squared: -0.009127
## F-statistic: 0.5025 on 3 and 162 DF, p-value: 0.6811tidy(test_fit2) # p value is small and significant## # A tibble: 4 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 22.4 3.92 5.71 0.0000000533
## 2 deep -0.0100 1.06 -0.00944 0.992
## 3 genderM 2.67 6.42 0.417 0.677
## 4 deep:genderM -0.408 1.71 -0.238 0.812summary(test_fit2)$r.squared # too low## [1] 0.009220341Basic data
# Female vs Male participants
csv_table_read %>%
ggplot(aes(x=gender)) +
geom_bar()
# age chart and gender per age
csv_table_read %>%
ggplot(aes(x= age, fill = gender)) +
geom_bar()
# age chart distribution per gender
csv_table_read %>%
ggplot(aes(x= age)) +
facet_grid(~gender) +
geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# age box plot distribution per gender
csv_table_read %>%
ggplot(aes(x= gender, y=age)) +
geom_boxplot()
# relationship and distribution between age, points, and gender
csv_table_read %>%
ggplot(aes(y = points, x = age, colour = gender)) +
geom_point() +
labs(title = "Distribution of points per age and gender")
# with this data we can observe the different age points that drives the mean up (vs the median).Gender analysis
Gender and points
# Distribution of the points per gender - histogram
csv_table_read %>%
ggplot(aes(x = points)) +
geom_histogram() +
facet_grid(~gender) +
labs(title = "Histogram of points by Gender")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
#Distribution of the points per gender - boxplot
csv_table_read %>%
ggplot(aes(y = points, x = gender, colour = gender)) +
geom_boxplot() +
labs(title = "Boxplot of points by Gender")
#QQ plot - points per gender
csv_table_read %>%
ggplot(aes(sample = points)) +
geom_qq() +
geom_qq_line(colour = "blue") +
facet_grid(~gender)
# mean points per gender - this is not significant
csv_table_read %>%
t.test(points ~ gender, data = .)##
## Welch Two Sample t-test
##
## data: points by gender
## t = -1.1832, df = 107.84, p-value = 0.2393
## alternative hypothesis: true difference in means between group F and group M is not equal to 0
## 95 percent confidence interval:
## -3.0896905 0.7799502
## sample estimates:
## mean in group F mean in group M
## 22.32727 23.48214Gender and attitude
# attitude vs gender
csv_table_read %>%
ggplot(aes(x=gender, y= attitude)) +
geom_boxplot()
# Type histogram
csv_table_read %>%
ggplot(aes(x = attitude)) +
geom_histogram() +
facet_grid(~ gender) +
labs(title = "Histogram of attitude by Gender")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# QQ plot: attitude per gender
csv_table_read %>%
ggplot(aes(sample = attitude)) +
geom_qq() +
geom_qq_line(colour = "blue") +
facet_grid(~gender)
# mean attitude per gender - This is significant and shows a difference between F and M on deep
csv_table_read %>%
t.test(attitude ~ gender, data = .)##
## Welch Two Sample t-test
##
## data: attitude by gender
## t = -4.0932, df = 122.66, p-value = 7.657e-05
## alternative hypothesis: true difference in means between group F and group M is not equal to 0
## 95 percent confidence interval:
## -0.6718635 -0.2338508
## sample estimates:
## mean in group F mean in group M
## 2.990000 3.442857Gender and deep learning approach:
# deep learning approach vs gender
# We could do that for all approach of learning
# Type histogram
csv_table_read %>%
ggplot(aes(x = deep)) +
geom_histogram() +
facet_grid(~ gender) +
labs(title = "Histogram of deep approach by Gender")## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
#Type boxplot
csv_table_read %>%
ggplot(aes(y = deep, x = gender, fill = gender)) +
geom_boxplot() +
labs(title = "Boxplot of deep Approach by Gender")
# QQ plot: deep per gender
csv_table_read %>%
ggplot(aes(sample = deep)) +
geom_qq() +
geom_qq_line(colour = "blue") +
facet_grid(~gender)
# mean deep per gender - This is quite significant and could show a correlation between the gender and the approach deep
csv_table_read %>%
t.test(deep ~ gender, data = .)##
## Welch Two Sample t-test
##
## data: deep by gender
## t = -0.72082, df = 101.32, p-value = 0.4727
## alternative hypothesis: true difference in means between group F and group M is not equal to 0
## 95 percent confidence interval:
## -0.2546963 0.1189279
## sample estimates:
## mean in group F mean in group M
## 3.656818 3.724702Age Analysis
Age and points
# does not seem to impact on points
csv_table_read %>%
ggplot(aes(x= age, y=points)) +
geom_point()+
facet_wrap(~ gender) +
geom_smooth(method = "lm")## `geom_smooth()` using formula = 'y ~ x'
Age and attitude
# does not seem to impact on attitude
csv_table_read %>%
ggplot(aes(x= age, y=attitude)) +
geom_point() +
facet_wrap(~ gender) +
geom_smooth(method = "lm")## `geom_smooth()` using formula = 'y ~ x'
Learning approach: deep analysis
deep and age
# deep learning approach vs age - no correlation
# We could do that for all approach of learning
csv_table_read %>%
ggplot(aes(x= age, y= deep)) +
geom_point() +
facet_wrap(~ gender) +
geom_smooth(method = "lm")## `geom_smooth()` using formula = 'y ~ x'
deep and points
# deep learning approach vs points - The deep approach seems to have a correlation with the number of points
csv_table_read %>%
ggplot(aes(x= deep, y=points)) +
geom_point() +
facet_wrap(~ gender) +
geom_smooth(method = "lm")## `geom_smooth()` using formula = 'y ~ x'
test_fit <- csv_table_read %>%
lm(points ~ deep * gender, data = .)
library(ggfortify)
summary(test_fit)##
## Call:
## lm(formula = points ~ deep * gender, data = .)
##
## Residuals:
## Min 1Q Median 3Q Max
## -15.3247 -3.3338 0.3369 4.6242 10.6787
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 22.36387 3.91828 5.708 5.33e-08 ***
## deep -0.01001 1.06032 -0.009 0.992
## genderM 2.67476 6.41719 0.417 0.677
## deep:genderM -0.40787 1.71487 -0.238 0.812
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.922 on 162 degrees of freedom
## Multiple R-squared: 0.00922, Adjusted R-squared: -0.009127
## F-statistic: 0.5025 on 3 and 162 DF, p-value: 0.6811# deep seems to have a significant impact on "points."Chapter 3
setwd("/Users/margot/Desktop/Desktop - MacBook Pro de MARGOT/Open data with R 2023/IODS-project")
library(broom)
library(ggplot2)
library(dplyr)
data <- read.csv("alc.csv",sep = ",", header = T)
class(data$high_use)## [1] "logical"class(data$famrel)## [1] "integer"class(data$goout)## [1] "integer"class(data$studytime)## [1] "integer"Question 3, 4 and 5
The purpose of your analysis is to study the relationships between high/low alcohol consumption and some of the other variables in the data. To do this, choose 4 interesting variables in the data and for each of them, present your personal hypothesis about their relationships with alcohol consumption. (0-1 point)
Hypothesis 1: students with a good family relationship have less chance to have a high consumption of alcohol
Comments on the below codes chunk: There is a negative correlation between family relationship and the consumption of alcohol. When the relationship is not good (from 1 to 5), there are more risk to drink lot of alcohol. Pvalue is 0,0204 which is representative(<0.05). The coefficient for the variable Famrel in this model is -0.2838. The odds ratio associated with a one-unit change in Famrel is approximately exp(-0.2838) = 0.753. For a one-unit decrease in Famrel, the odds of ‘high_use’ decrease by about 1 - 0.753 = 24.7%. With 95% confidence, the true value of the odds ratio for the intercept lies between approximately 0.49 and 3.35. For the famrel, the odds ratio lies between 0.59 and 0.95, since it is less than 1, there is a statistical significance, with a negative effect from the variable famrel to the high_use. The null deviance is bigger than the residual deviance of the model. It shows that the model explains a part of the variable high-use.
# I add the family = binomial as we do the analyze on a binary variable (True and False)
logistic_model_family_alcohol <- glm(high_use ~ famrel, data = data, family = binomial)
summary(logistic_model_family_alcohol)##
## Call:
## glm(formula = high_use ~ famrel, family = binomial, data = data)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.2573 0.4853 0.530 0.5960
## famrel -0.2838 0.1224 -2.318 0.0204 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 446.67 on 368 degrees of freedom
## AIC: 450.67
##
## Number of Fisher Scoring iterations: 4# get the coef
coef(logistic_model_family_alcohol) %>% exp()## (Intercept) famrel
## 1.2934265 0.7528865# odd of being in high use - exponential of the coef
exp(coef(logistic_model_family_alcohol))## (Intercept) famrel
## 1.2934265 0.7528865# get the intervals
confint(logistic_model_family_alcohol) %>% exp()## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) 0.4961949 3.3547129
## famrel 0.5911258 0.9570208# get the variance, BIC and AIC
logistic_model_family_alcohol %>%
glance()## # A tibble: 1 × 8
## null.deviance df.null logLik AIC BIC deviance df.residual nobs
## <dbl> <int> <dbl> <dbl> <dbl> <dbl> <int> <int>
## 1 452. 369 -223. 451. 458. 447. 368 370# If I want to check the impact of high use to family relationship -> famrel is ordinal so I use the ordinal regression model
library(ordinal)##
## Attaching package: 'ordinal'## The following object is masked from 'package:dplyr':
##
## sliceordinal_model_family_alcohol <- clm(factor(famrel) ~ high_use, data = data)
summary(ordinal_model_family_alcohol)## formula: factor(famrel) ~ high_use
## data: data
##
## link threshold nobs logLik AIC niter max.grad cond.H
## logit flexible 370 -454.70 919.40 5(0) 3.77e-08 2.0e+01
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## high_useTRUE -0.5411 0.2139 -2.529 0.0114 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Threshold coefficients:
## Estimate Std. Error z value
## 1|2 -4.0079 0.3672 -10.914
## 2|3 -2.7746 0.2194 -12.645
## 3|4 -1.3118 0.1422 -9.222
## 4|5 0.8468 0.1300 6.516# This box plot confirms our hypothesis and conclusion. We can see that the mean and median of the family relationship quality is very different whether the students drink a lot or not.
data %>%
ggplot(aes(x= factor(high_use), y= famrel)) +
geom_boxplot() +
labs(title = "Quality of family relationship vs the alcohol consumption",
x = "High consumption",
y = "quality of family relationship")
data %>%
ggplot(aes(x = as.factor(famrel), fill = as.factor(high_use))) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Family relationship and Alcohol Consumption",
x = "Family relation score from 1 to 5",
y = "Proportion of students",
fill = "High Use of alcohol: True of False"
)
# there are less score 5 and way more score 2 among the students who drink a lot
data %>%
ggplot(aes(x = as.factor(high_use), fill = as.factor(famrel))) +
geom_bar(position = "fill") +
labs(
title = "Family relation score distribution by Alcohol Consumption",
x = "High Use of alcohol: True of False",
y = "Proportion of students",
fill = "Family relation score from 1 to 5"
)
Hypothesis 2: students who go more out with their friends have more risk to have a high consumption of alcohol
Comments on the below codes chunk: There is a a positive relationship between going out and the consumption of alcohol and a significant p value (very small value), which confirms the hypothesis. The more students go out the more risks they have to have a high consumption in alcohol.
The coefficient for goout is 0.7563.The odds ratio associated with a one-unit change in Goout is approximately exp(0.7563) = 2.13.For a one-unit increase in Goout, the odds of ‘high_use’ increase by about 2.13 - 1 = 1.13, which is about 113% increase.
With 95% confidence, the true value of the odds ratio for the intercept lies between approximately 0.015 and 0.078. For the goout, the odds ratio lies between 1.71 and 2.66, since it is > 1, there is a statistical significance with a positive effect of this variable on the high_use.
The null deviance is way bigger than the residual deviance of the model. It shows that the model explains the variable high-use, even better than in the hypothesis 1. The AIC of this model is even smaller than the previous model too, which confirms my conclusion.
logistic_model_goout_alcohol <- glm(high_use ~ goout, data = data, family = binomial)
summary(logistic_model_goout_alcohol)##
## Call:
## glm(formula = high_use ~ goout, family = binomial, data = data)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.3368 0.4188 -7.968 1.62e-15 ***
## goout 0.7563 0.1165 6.494 8.34e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 403.05 on 368 degrees of freedom
## AIC: 407.05
##
## Number of Fisher Scoring iterations: 4tidy(logistic_model_goout_alcohol)## # A tibble: 2 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -3.34 0.419 -7.97 1.62e-15
## 2 goout 0.756 0.116 6.49 8.34e-11# get the coef
coef(logistic_model_goout_alcohol) %>% exp()## (Intercept) goout
## 0.03554946 2.13048020# get the exp of the coef
exp(coef(logistic_model_goout_alcohol))## (Intercept) goout
## 0.03554946 2.13048020# get the intervals
confint(logistic_model_goout_alcohol) %>% exp()## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) 0.01515176 0.07850224
## goout 1.70566958 2.69518608# plots
data %>%
ggplot(aes(x= factor(high_use), y= goout)) +
geom_boxplot() +
labs(title = "Going out with friends vs the alcohol consumption",
x = "High consumption",
y = "go out with friends (from 1 to 5)")
# the students who are drinking more are more numerous among the students who gave the score 4 and 5 for the goout variable.
data %>%
ggplot(aes(x = factor(goout), fill = factor(high_use))) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Going Out Behavior and Alcohol Consumption",
x = "going out with friends from 1 to 5",
y = "Proportion of students",
fill = "High Use of alcohol: True of False"
)
# there are more score 1, 2 and 3 among the students who do not drink a lot
data %>%
ggplot(aes(x = factor(high_use), fill = factor(goout))) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Alcohol Consumption and going out behavior",
x = "High Use of alcohol: True of False",
y = "Proportion of students",
fill = "going out with friends from 1 to 5"
)
Hypothesis 3: students who study a lot have less risk to have a high consumption of alcohol
Comments about the below code chunk: The p value is very small and estimate is negative. We can conclude there is a negative relation between the study time and the alcohol consumption. When the study time increases the odds of high alcohol consumption decreases.
The coefficient for Studytime is -0.6208. The odds ratio associated with a one-unit change in Studytime is approximately exp(-0.6208) = 0.538. For a one-unit decrease in Studytime, the odds of ‘high_use’ decrease by about 1 - 0.538 = 46.2%.
With 95% confidence, the true value of the odds ratio for the intercept lies between approximately 0.79 and 2.67. For the study time, the odds ratio lies between 0.39 and 0.72, since it is < 1, there is a statistical significance with a negative effect of this variable on the high_use.
The residual deviance is smaller than the null deviance, it means the model explains the variable high use. The AIC is smaller than the first model but not as small as the second model.
logistic_model_studytime_alcohol <- glm(high_use ~ studytime, data = data, family = binomial)
summary(logistic_model_studytime_alcohol)##
## Call:
## glm(formula = high_use ~ studytime, family = binomial, data = data)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 0.3649 0.3102 1.176 0.239
## studytime -0.6208 0.1542 -4.027 5.66e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 433.82 on 368 degrees of freedom
## AIC: 437.82
##
## Number of Fisher Scoring iterations: 4tidy(logistic_model_studytime_alcohol)## # A tibble: 2 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) 0.365 0.310 1.18 0.239
## 2 studytime -0.621 0.154 -4.03 0.0000566# get the coef
coef(logistic_model_studytime_alcohol) %>% exp()## (Intercept) studytime
## 1.4403849 0.5375316#Get the odds - exponential of coefficient:
exp(coef(logistic_model_studytime_alcohol))## (Intercept) studytime
## 1.4403849 0.5375316# get the intervals
confint(logistic_model_studytime_alcohol) %>% exp()## Waiting for profiling to be done...## 2.5 % 97.5 %
## (Intercept) 0.7881099 2.6654451
## studytime 0.3933727 0.7208174# This plot shows the difference in median and mean for the students who drink a lot or not
data %>%
ggplot(aes(x= factor(high_use), y= studytime)) +
geom_boxplot() +
labs(title = "Study time vs the alcohol consumption",
x = "High consumption",
y = "Study time (from 1 to 4)")
# way more students among students who score they study time around 1 and 2
data %>%
ggplot(aes(x = factor(studytime), fill = factor(high_use))) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Study time Behavior and Alcohol Consumption",
x = "Study time from 1 to 4",
y = "Proportion of students",
fill = "High Use of alcohol: True of False"
)
# way more score 1 and way less 3 and 4 among students who drink a lot.
data %>%
ggplot(aes(x = factor(high_use), fill = factor(studytime))) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Alcohol Consumption and Study time behavior",
x = "High Use of alcohol: True of False",
y = "Proportion of students",
fill = "Study time from 1 to 4"
)
Hypothesis 4: students who chose the school because of its reputation has less risk to have a high consumption of alcohol
Comments about the below code: The p value is 0,022 which tells that there is a correlation between the reason to choose a school and the consumption of alcohol.
We don’t know with the test which reason is the most chosen by the students who drink a lot of alcohol vs the ones who don’t. Some plots can help us identify it (check below).
# Creating a contingency table
chi_table_alcohol_reasons <- table((data$high_use), as.factor(data$reason))
# Performing the chi-squared test
chi_square_test <- chisq.test(chi_table_alcohol_reasons )
chi_square_test##
## Pearson's Chi-squared test
##
## data: chi_table_alcohol_reasons
## X-squared = 9.563, df = 3, p-value = 0.02267# Summary of the test
summary(chi_square_test)## Length Class Mode
## statistic 1 -none- numeric
## parameter 1 -none- numeric
## p.value 1 -none- numeric
## method 1 -none- character
## data.name 1 -none- character
## observed 8 table numeric
## expected 8 -none- numeric
## residuals 8 table numeric
## stdres 8 table numeric# This show that students who don't drink a lot have chosen their school more for its reputation than the one who drink a lot. For the ones who drink, the reasons home and other are way more important than for the students who don't drink too much. The reason courses seem to be equally important for them.
data %>%
ggplot(aes(x = high_use, fill = reason)) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Reason to chose their school and Alcohol Consumption",
x = "High Consumption of a",
y = "Proportion of students",
fill = "High Use of alcohol: True of False"
)
data %>%
ggplot(aes(x = reason, fill = high_use)) +
geom_bar(position = "fill") +
labs(
title = "Relationship Between Alcohol Consumption and Reason to chose their school",
x = "High Use of alcohol: True of False",
y = "Proportion of students",
fill = "Reason to choose the school"
)
Other models to test:
By adding each variable together, we can see that the variation in high use is better explained (this is because the deviance is decreasing from 452 to 379).
All three variables (goout, famrel, and studytime) have statistical significance in predicting alcohol consumption (high_use).
logistic_model_3var <- glm(high_use ~ goout + famrel + studytime, data = data, family = binomial)
logistic_model_3var_2 <- glm(high_use ~ goout * famrel * studytime, data = data, family = binomial)
summary(logistic_model_3var)##
## Call:
## glm(formula = high_use ~ goout + famrel + studytime, family = binomial,
## data = data)
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.7286 0.6941 -1.050 0.29381
## goout 0.7777 0.1203 6.467 9.99e-11 ***
## famrel -0.3777 0.1367 -2.763 0.00573 **
## studytime -0.6139 0.1672 -3.673 0.00024 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 452.04 on 369 degrees of freedom
## Residual deviance: 379.74 on 366 degrees of freedom
## AIC: 387.74
##
## Number of Fisher Scoring iterations: 4tidy(logistic_model_3var)## # A tibble: 4 × 5
## term estimate std.error statistic p.value
## <chr> <dbl> <dbl> <dbl> <dbl>
## 1 (Intercept) -0.729 0.694 -1.05 2.94e- 1
## 2 goout 0.778 0.120 6.47 9.99e-11
## 3 famrel -0.378 0.137 -2.76 5.73e- 3
## 4 studytime -0.614 0.167 -3.67 2.40e- 4# Anova conclusion: The p-value associated with the difference in deviance (Pr(>Chi)) is 0.1362, suggesting that the difference in fit between the two models is not statistically significant
anova(logistic_model_3var, logistic_model_3var_2, test="LRT")## Analysis of Deviance Table
##
## Model 1: high_use ~ goout + famrel + studytime
## Model 2: high_use ~ goout * famrel * studytime
## Resid. Df Resid. Dev Df Deviance Pr(>Chi)
## 1 366 379.74
## 2 362 372.74 4 6.995 0.1362# when comparing with for instanc: logistic_model_family_alcohol, the residual deviance of logistic_model_3var is better and the p is significant.
anova(logistic_model_3var, logistic_model_family_alcohol, test="LRT")## Analysis of Deviance Table
##
## Model 1: high_use ~ goout + famrel + studytime
## Model 2: high_use ~ famrel
## Resid. Df Resid. Dev Df Deviance Pr(>Chi)
## 1 366 379.74
## 2 368 446.67 -2 -66.934 2.921e-15 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# best model is the logistic_model_3var according to BIC and AIC, because metrics are smaller for this model
AIC(logistic_model_3var, logistic_model_3var_2)## df AIC
## logistic_model_3var 4 387.7372
## logistic_model_3var_2 8 388.7422BIC(logistic_model_3var, logistic_model_3var_2)## df BIC
## logistic_model_3var 4 403.3912
## logistic_model_3var_2 8 420.0502question 6: Model 1 (high_use ~ goout + famrel + studytime)
Using the variables which, according to your logistic regression model, had a statistical relationship with high/low alcohol consumption, explore the predictive power of you model. Provide a 2x2 cross tabulation of predictions versus the actual values and optionally display a graphic visualizing both the actual values and the predictions. Compute the total proportion of inaccurately classified individuals (= the training error) and comment on all the results. Compare the performance of the model with performance achieved by some simple guessing strategy.
# prediction based on the model: logistic_model_3var
predictions <- predict(logistic_model_3var, type = "response")
#Create the confusion matrix:
table(data$high_use, predictions > 0.5)##
## FALSE TRUE
## FALSE 235 24
## TRUE 63 48# correct are False false + true true = 235 + 48 = 293
# wrong are False True and True False = 24 + 63 = 87
# Probability of miss classifications = 87/(293+87) = 23%
# odds of miss classification = 87/293 = 30%
# precision: correctly positive among all positives predictions = 0.33%
precision <- 24 / (24 + 48)
# recall: correctly positive among all actual positives = 0.28%
recall <- 24 / (24 + 63)
# F1 score: 0.30%
2 * (precision * recall) / (precision + recall)## [1] 0.3018868 # Accuracy of always predicting the majority class: if the majority class is "FALSE," then predicting "FALSE" for every instance would result in a correct prediction rate of 70%
mean(data$high_use == FALSE)## [1] 0.7# Calculate the total proportion of errors is 23.5%. This means my model is a bit better than the majority class which is 70% (30% of non-accuracy).
mean(data$high_use != (predictions > 0.5))## [1] 0.2351351# to conclude, my model is a bit better than using the prediction based on the majority class. But it should be improved for better accuracy!Bonus:
I had to look at chat gpt fully for this one. I did not know the method. I now understand this method splits the data in 10 data samples and will use 9 of them to predict the last one, rotating to test them all. In that case, the average of prediction for the different tests is 74.9% which is better than the 70% (majority class).
library(caret)## Loading required package: lattice##
## Attaching package: 'caret'## The following object is masked from 'package:purrr':
##
## liftlibrary(lattice)
# Define the model formula
formula <- factor(high_use) ~ goout + famrel + studytime
# Define the number of folds for cross-validation
num_folds <- 10
# Create a training control with 10-fold cross-validation
train_control <- trainControl(method = "cv", number = num_folds)
# Train the model using 10-fold cross-validation
model <- train(formula, data = data, method = "glm", family = binomial, trControl = train_control)
# Get cross-validated results
results <- model$resample
str(results)## 'data.frame': 10 obs. of 3 variables:
## $ Accuracy: num 0.757 0.763 0.703 0.73 0.73 ...
## $ Kappa : num 0.33 0.412 0.228 0.277 0.353 ...
## $ Resample: chr "Fold01" "Fold02" "Fold03" "Fold04" ...# Calculate mean prediction error
mean(results$Accuracy)## [1] 0.7622017Question 1, 2 and 3
dataset description:
The Boston research seems to look for correlations between the standard of living, housing quality, and other criteria such as the urban environment, accessibility, and demographics within the Boston area.
Each row represents data for a specific town and has information about various factors such as crime rate, proportion of non-retail business acres, property tax rates, pupil-teacher ratios, etc.
This data frame contains the following columns:
crim - per capita crime rate by town.
zn - proportion of residential land zoned for lots over 25,000 sq.ft.
indus - proportion of non-retail business acres per town.
chas - Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).
nox - nitrogen oxides concentration (parts per 10 million).
rm - average number of rooms per dwelling.
age - proportion of owner-occupied units built prior to 1940.
dis - weighted mean of distances to five Boston employment centres.
rad - index of accessibility to radial highways.
tax - full-value property-tax rate per $10,000.
ptratio - pupil-teacher ratio by town.
black - 1000(Bk−0.63) square Bk is the proportion of blacks by town.
lstat - lower status of the population (percent).
medv - median value of owner-occupied homes in $1000s.
Data analyzes:
Explore the structure and the dimensions of the dataset. Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, the distributions of the variables and the relationships between them.
library(MASS)##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## selectlibrary(dplyr)
library(ggplot2)
library(GGally)
library(tidyr)
setwd("/Users/margot/Desktop/Desktop - MacBook Pro de MARGOT/Open data with R 2023/IODS-project")
Boston <- Boston
# explore the mean, median, min, max per variable
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00# 506 entries with 14 variables
dim(Boston)## [1] 506 14# Data type: all num or int
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...# let's create points plot for each variable. -> not easy to read as too many variables
pairs(Boston)
# Let's analyze the histogram for each variable:
gather(Boston) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") +
geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# Let's compute a correlation matrix
matrix_correlation_var <- cor(Boston)
# Visualize correlation matrix as a heatmap
library(reshape2)##
## Attaching package: 'reshape2'## The following object is masked from 'package:tidyr':
##
## smithsggplot(data = melt(matrix_correlation_var), aes(Var1, Var2, fill = value)) +
geom_tile() +
scale_fill_gradient2(low = "blue", high = "red", mid = "white", midpoint = 0)
according to the matrix - Negative relationships between certain variables:
lstat and medv - lower status of population and the median price of homes owned by occupants
lstat and rm - lower status of population and the average of room numbers per home
tax and medv - the percentage at which a property is taxed based on its assessed value and the the median price of homes owned by occupants
dis and lstat - the weighted average distance from each house to these employment centers and the lower status of population
dis and age - the weighted average distance from each house to these employment centers and the percentage or fraction of homes that are owner-occupied and were constructed before the year 1940.
dis and nox - the weighted average distance from each house to these employment centers and nitrogen oxides concentration.
dis and indus - the weighted average distance from each house to these employment centers and the proportion of non-retail business acres per town (manufacturing facilities, industrial parks, office spaces, warehouses etc).
tax and dis - the percentage at which a property is taxed based on its assessed value and the weighted average distance from each house to these employment centers
zn and age - the percentage or fraction of land within residential areas that is for individual lots with a minimum size of over 25,000 square feet and the percentage or fraction of homes that are owner-occupied and were constructed before the year 1940
zn and nox - the percentage or fraction of land within residential areas that is for individual lots with a minimum size of over 25,000 square feet and the nitrogen oxides concentration.
zn and indus - the percentage or fraction of land within residential areas that is for individual lots with a minimum size of over 25,000 square feet and the proportion of non-retail business acres per town (manufacturing facilities, industrial parks, office spaces, warehouses etc).
Positive relationships between variables
medv and rm - the median price of homes owned by occupants and the average of room numbers per home
tax and indus - the percentage at which a property is taxed based on its assessed value and the proportion of non-retail business acres per town.
tax and nox - the percentage at which a property is taxed based on its assessed value and the nitrogen oxides concentration.
age and indus - the percentage or fraction of homes that are owner-occupied and were constructed before the year 1940 and the proportion of non-retail business acres per town.
age and nox- the percentage or fraction of homes that are owner-occupied and were constructed before the year 1940 and the nitrogen oxides concentration.
nox and indus - the nitrogen oxides concentration and the proportion of non-retail business acres per town.
–> tests in my model: sltat, medv, rm, tax, dis, age, nox, indus, zn
Test some models based on the dataset and observed correlation
library(car)## Loading required package: carData##
## Attaching package: 'car'## The following object is masked from 'package:dplyr':
##
## recode## The following object is masked from 'package:purrr':
##
## somemodel1 <- glm(lstat ~ medv + dis + rm,data=Boston)
model2 <- glm(medv ~ rm + tax + lstat ,data=Boston)
# all the VIF are between 1 and 2.1 which is ok. It suggest a low multicollinearity and imply that the variance of the estimated coefficients is not significantly inflated due to collinearity
vif(model1)## medv dis rm
## 1.982257 1.068810 1.940168vif(model2)## rm tax lstat
## 1.610953 1.426005 2.092901#let’s calculate the corresponding ODDS ratios and confidence intervals (95%):
OR1 <- coef(model1) %>% exp
OR2 <- coef(model2) %>% exp
CI1 <- confint(model1) %>% exp## Waiting for profiling to be done...CI2 <- confint(model2) %>% exp## Waiting for profiling to be done...# the confidence interval for an odds ratio doesn't span 1 = there's a statistically significant effect for both models
cbind(OR1, CI1) ## OR1 2.5 % 97.5 %
## (Intercept) 1.764803e+16 4.023272e+14 7.741285e+17
## medv 6.606608e-01 6.250484e-01 6.983022e-01
## dis 3.292580e-01 2.756488e-01 3.932934e-01
## rm 1.682751e-01 8.210665e-02 3.448749e-01cbind(OR2, CI2)## OR2 2.5 % 97.5 %
## (Intercept) 0.6073487 0.001289619 286.0319984
## rm 181.1868455 76.546116788 428.8744403
## tax 0.9935198 0.990167804 0.9968832
## lstat 0.5754723 0.522466602 0.6338557# the residual deviance is way smaller than the null deviance. It indicates a reasonably good fit of the model to the data.
summary(model1)##
## Call:
## glm(formula = lstat ~ medv + dis + rm, data = Boston)
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 37.40940 1.92918 19.391 < 2e-16 ***
## medv -0.41451 0.02827 -14.662 < 2e-16 ***
## dis -1.11091 0.09067 -12.252 < 2e-16 ***
## rm -1.78215 0.36612 -4.868 1.51e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 17.22406)
##
## Null deviance: 25752.4 on 505 degrees of freedom
## Residual deviance: 8646.5 on 502 degrees of freedom
## AIC: 2882.2
##
## Number of Fisher Scoring iterations: 2# medv and rm also influences each other, so let's modify the model a bit
model11 <- glm(lstat ~ medv + dis + rm + rm * medv ,data=Boston)
summary(model11)##
## Call:
## glm(formula = lstat ~ medv + dis + rm + rm * medv, data = Boston)
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 74.41052 3.64466 20.42 <2e-16 ***
## medv -1.94316 0.13517 -14.38 <2e-16 ***
## dis -0.89569 0.08285 -10.81 <2e-16 ***
## rm -7.55541 0.59817 -12.63 <2e-16 ***
## medv:rm 0.22526 0.01957 11.51 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 13.64918)
##
## Null deviance: 25752.4 on 505 degrees of freedom
## Residual deviance: 6838.2 on 501 degrees of freedom
## AIC: 2765.5
##
## Number of Fisher Scoring iterations: 2# same for this one, residual deviance is way smaller than the null deviance.
summary(model2)##
## Call:
## glm(formula = medv ~ rm + tax + lstat, data = Boston)
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -0.498652 3.140239 -0.159 0.873895
## rm 5.199529 0.439618 11.827 < 2e-16 ***
## tax -0.006501 0.001724 -3.770 0.000182 ***
## lstat -0.552564 0.049302 -11.208 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for gaussian family taken to be 29.90866)
##
## Null deviance: 42716 on 505 degrees of freedom
## Residual deviance: 15014 on 502 degrees of freedom
## AIC: 3161.4
##
## Number of Fisher Scoring iterations: 2Question 4
Standardizing the data. Variables often have different scales and units, making direct comparisons challenging. Standardization brings all variables to a common scale, allowing for fair comparisons between different variables. It makes the distribution of each variable more consistent, with a mean of 0 and a standard deviation of 1. This normalization aids in interpreting coefficients and comparing the relative importance of different predictors in regression. Standardizing ensures that each variable contributes equally to these techniques, preventing one variable from dominating the analysis due to its scale. It allows easier comparison of the magnitude of the effect of each variable on the outcome. Finally it can mitigate issues related to multicollinearity in regression analysis by putting variables on a similar scale, reducing the impact of differing scales on regression coefficients.
# select numerical values
Boston_numeric_cols <- Boston[, sapply(Boston, is.numeric)]
# The scale() to standardize and transform the data to have a mean of 0 and a standard deviation of 1.
scaled_boston <- scale(Boston_numeric_cols)
# convert to a data frame
scaled_table_boston <- as.data.frame(scaled_boston)
# how did the data change? Mean is now 0 so it has worked;
summary(scaled_table_boston)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865# use the cut function to create categorical variables based on intervals or breaks in a numerical variable. We do this process for the crim variable from 0 to 0.25 to 0.50 to 0.75 to 1 (quantiles). Add labels for each category.
# include.lowest = TRUE is to ensure there is no NA category.
quantiles <- quantile(Boston$crim, probs = seq(0, 1, by = 0.25), na.rm = TRUE)
interval_labels <- c("premier_quantil", "second_quantil", "third_quantil", "fourth_quantil")
scaled_table_boston$quantiles_crime <- cut(Boston$crim, quantiles, labels= interval_labels,include.lowest = TRUE)
# Notes: Quantiles derived from numeric values can be considered categorical or continuous. When quantiles represent discrete categories or bins that divide a continuous variable into distinct groups, they are treated as categorical (e. small, medium, big). If quantiles represent numeric values that indicate the position or value relative to the distribution of a continuous variable (e.g., the 25th percentile, median, 75th percentile), they are considered continuous.
# drop the former column crim and create a new table
Boston_new <- scaled_table_boston
# For some reasons, it does not knit via the index file when I have this "select" function. However, the Knit works normally when I do it from the chapter 4 with or without this function. Therefore I have to remove it to be able to knit it via the index.rmd.
# In order to remove the crim, i do as below.
#%>%select(-crim, everything())
# We need 80% of the rows from total rows
train_size <- round(0.8 * nrow(Boston_new))
# Select a sample randomly among the dataset 80%
train_set <- sample(seq_len(nrow(Boston_new)), size = train_size)
# Create training and testing subsets
train_data <- Boston_new[train_set, ]
test_data <- Boston_new[-train_set, ]Question 5:
Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot.
Linear Discriminant Analysis (LDA) can be quite useful in various real-life scenarios where classification or dimensionality reduction is required such as: Marketing and Customer Segmentation, product quality control, credit scoring.
It is ideal when we have multiple variables that can help us categorize each entry in one specific group where all others will have similar mean, and variance. The goal is to predict the categorical outcome or class based on a set of predictors.
LDA transforms the original variables into a smaller set of new variables, known as linear discriminants. These new variables (linear discriminants) are created in a way that maximizes the separation between different categories or classes in the data.By using the information captured in the linear discriminants, LDA helps in accurately assigning new observations (data points) to their respective categories or classes. LDA takes variables, combines them in a smart way to create new variables that are helpful for understanding differences between groups or categories, and then uses this understanding to categorize or classify new data based on what it has learned. This makes it useful for both simplifying data and making predictions about categories in that data.
LDA calculates the mean and variance for each predictor within each class or category in the target variable. It finds coefficients for these combinations by considering differences in means between classes and assumes equal variances for each predictor within classes. It assumes that the data approximately follows a normal distribution.
# check the impact of each variable on a categorical variable (quantiles_crime)
# each quantiles are approximately equal 25%
# LD1 explains 96% of the model. When LD1 captures 96% of the variance, it suggests that LD1 effectively separates the classes or groups in the data based on their mean differences and variance. Within LD1, data points are grouped together in a way that maximizes the separation between classes while minimizing the variation within each class.
# Positive coefficients indicate that higher values of that variable contribute to a higher score in that particular discriminant, while negative coefficients suggest the opposite. The larger the magnitude of the coefficient, the greater the impact of that variable on the discriminant.
library(MASS)
# Fit LDA on the train set - LDA is a statistical method used for classification. It fits an LDA model using the train_data, where quantiles_crime is the target variable to predict based on the other variables
lda_model <- lda(quantiles_crime ~ ., data = train_data)
# Predict on the test set by using the function predict to apply the model created from the train_data to test_data -> the idea here is to predict the class labels for this test_data
lda_pred <- predict(lda_model, test_data)$class # class extract the predicted class labels
# actual crime quantiles
actual_crime_categories <- test_data$quantiles_crime
# Extract LD scores from the LDA model's predictions
lda_scores <- predict(lda_model, test_data)$x # x accesses the matrix of posterior probabilities or scores associated with each class for the observations in the test_data.
# Create a dataframe with LD scores from first 2 columns of the lda_score and the predicted classes. Combining LD1, LD2, and the predicted classes for visualization. In many cases, visualizing beyond LD1 and LD2 might become complex to display effectively in two-dimensional plots. LD1 and LD2 are typically chosen for visualization as they capture the most discrimination power while allowing for a clearer visualization on a 2D plot.
plot_data <- data.frame(LD1 = lda_scores[, 1], LD2 = lda_scores[, 2], prediction_crime_quantile = as.factor(lda_pred))
plot_data## LD1 LD2 prediction_crime_quantile
## 3 -3.220127515 -0.842606699 second_quantil
## 7 -1.983476161 -0.484841597 second_quantil
## 8 -1.385730986 -1.376476517 third_quantil
## 12 -1.891100414 -0.693218970 second_quantil
## 15 -1.958469303 -0.690055230 second_quantil
## 17 -2.479946070 -0.005556002 second_quantil
## 24 -1.607577196 -0.984618924 third_quantil
## 36 -2.094443430 0.137798834 second_quantil
## 39 -2.303340863 0.292891577 second_quantil
## 46 -3.309354900 0.302657621 second_quantil
## 47 -3.193872611 0.125648576 second_quantil
## 49 -2.168842909 -1.175718288 second_quantil
## 51 -2.807075046 0.512875355 second_quantil
## 53 -3.219419597 0.862565430 premier_quantil
## 57 -3.608059610 2.424840887 premier_quantil
## 59 -1.497461279 1.173514629 second_quantil
## 65 -2.903939775 -0.429975032 second_quantil
## 66 -3.333080601 3.210585904 premier_quantil
## 74 -3.292685041 0.635007696 second_quantil
## 75 -2.916890849 0.634136244 second_quantil
## 79 -2.316857615 -0.223919792 second_quantil
## 86 -3.073630573 -0.037578054 second_quantil
## 102 -1.805210668 -0.393888146 second_quantil
## 111 -1.884219814 -0.096201210 second_quantil
## 113 -1.173015218 -0.730861567 second_quantil
## 115 -1.470004381 -0.416205832 second_quantil
## 125 -2.212090876 -1.770975443 third_quantil
## 126 -2.316744514 -1.681729830 third_quantil
## 127 -1.957448622 -1.929169347 third_quantil
## 132 -1.373927612 -1.736843574 third_quantil
## 135 -1.077157489 -1.798155344 third_quantil
## 146 0.194144281 -3.457582986 third_quantil
## 149 0.110132008 -3.371916033 third_quantil
## 157 0.004101181 -2.933539829 third_quantil
## 160 -0.463299189 -3.107242623 third_quantil
## 180 -2.786201112 -0.300729426 premier_quantil
## 184 -2.462343806 -0.581426034 second_quantil
## 189 -2.725988089 1.818370362 premier_quantil
## 201 -3.406914766 2.996048976 premier_quantil
## 207 -2.484502250 -0.399695028 second_quantil
## 210 -1.985591738 -1.220710698 second_quantil
## 225 -0.788632868 -0.792314991 third_quantil
## 228 -0.910551638 -0.309595946 third_quantil
## 229 -1.286212844 -0.085146771 third_quantil
## 230 -1.524996065 0.601160633 second_quantil
## 233 -0.950638840 -0.739976319 third_quantil
## 235 -1.312602277 -0.280505787 second_quantil
## 242 -2.005060803 0.907556303 second_quantil
## 246 -1.248486620 0.274945765 second_quantil
## 247 -1.778741461 0.775152873 second_quantil
## 253 -2.173361128 0.878523999 second_quantil
## 257 -3.137796095 2.443217413 premier_quantil
## 258 -1.263386496 -1.516328484 third_quantil
## 263 -1.179623410 -1.618210891 third_quantil
## 267 -1.314960941 -0.942571385 third_quantil
## 272 -3.362290408 1.077972858 premier_quantil
## 281 -2.216934863 -0.113127199 premier_quantil
## 284 -4.266165910 2.039121323 premier_quantil
## 292 -2.800856645 2.782884644 premier_quantil
## 294 -3.219617026 0.052001086 second_quantil
## 299 -3.031942369 2.718750193 premier_quantil
## 307 -1.370913097 1.135387153 premier_quantil
## 310 -2.145521259 -0.566737142 second_quantil
## 313 -1.990874420 -0.731240248 second_quantil
## 314 -2.137409876 -0.688564078 second_quantil
## 315 -2.023344458 -0.980821081 third_quantil
## 316 -2.159838060 -0.601562588 second_quantil
## 317 -1.877548714 -1.023462339 second_quantil
## 325 -2.344152148 0.073420402 second_quantil
## 331 -2.992697602 0.122461697 second_quantil
## 334 -2.313262108 -0.070951837 second_quantil
## 336 -2.297328829 0.058652234 second_quantil
## 341 -2.088716945 0.020823934 second_quantil
## 342 -3.826662804 0.252445049 premier_quantil
## 343 -3.719395499 -0.837889797 second_quantil
## 352 -2.897968657 1.444438550 premier_quantil
## 353 -3.107461945 1.897953412 premier_quantil
## 354 -2.506180478 2.185986928 premier_quantil
## 360 5.855922201 -0.391016936 fourth_quantil
## 368 6.341239856 0.568791183 fourth_quantil
## 372 6.262260097 -0.499180443 fourth_quantil
## 385 6.483799737 0.350698960 fourth_quantil
## 386 6.185929979 0.233661289 fourth_quantil
## 388 6.287938553 0.391080220 fourth_quantil
## 389 6.315507724 0.060243799 fourth_quantil
## 390 5.841004455 0.223041462 fourth_quantil
## 395 5.719489774 0.445362246 fourth_quantil
## 403 5.761207712 0.159549005 fourth_quantil
## 404 6.004886422 0.716584744 fourth_quantil
## 409 5.812791125 0.385372873 fourth_quantil
## 413 6.833141928 0.134702175 fourth_quantil
## 422 5.812900249 0.090319553 fourth_quantil
## 429 5.862910942 0.313407081 fourth_quantil
## 436 6.395175936 -0.447766865 fourth_quantil
## 444 5.980631885 -0.259357534 fourth_quantil
## 449 5.818946101 -0.027371392 fourth_quantil
## 451 6.153922581 -0.221741397 fourth_quantil
## 455 6.299353509 -0.334921305 fourth_quantil
## 456 6.048323719 -0.191491437 fourth_quantil
## 458 6.145544626 -0.018600119 fourth_quantil
## 490 -1.251907115 -1.909555446 second_quantil# Create a scatterplot of LD1 and LD2
plot_LDA <- ggplot(plot_data, aes(x = LD1, y = LD2, color = prediction_crime_quantile)) +
geom_point() +
labs(title = "LDA Biplot with Predicted Crime Quantiles")
plot_LDA
# adding real values - comparison of actual vs predicted values in test_data
realVsPred_plot <- plot_LDA +
geom_point(aes(color = actual_crime_categories), size = 4, alpha = 0.1) +
labs(color = "Real Quantiles of Crime")
realVsPred_plot
# the accuracy of predictions using test data
accuracy <- mean(lda_pred == test_data$quantiles_crime)
print(paste("Accuracy of LDA model on test data:", round(accuracy * 100, 2), "%"))## [1] "Accuracy of LDA model on test data: 73.27 %"Question 6:
Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results. (0-3 points)
# save the crime data:
actual_crime_categories <- test_data$quantiles_crime
# Remove the categorical crime variable from the test dataset
test_data_without_crime <- subset(test_data, select = -c(quantiles_crime))
# get the classes based on the model - this was done earlier with lda_pred. so I am a bit confused.
lda_pred_test <- predict(lda_model, test_data_without_crime)$class
# get the table with the prediction vs actual - results are same between the 2 ways since I did 2 times the same steps. I might have missed something in the requests.
cross_tab <- table(Predicted = lda_pred_test, Actual = actual_crime_categories)
cross_tab## Actual
## Predicted premier_quantil second_quantil third_quantil fourth_quantil
## premier_quantil 15 2 0 0
## second_quantil 11 21 11 0
## third_quantil 0 3 16 0
## fourth_quantil 0 0 0 22cross_table <- table(Predicted = lda_pred, Actual = actual_crime_categories)
cross_table## Actual
## Predicted premier_quantil second_quantil third_quantil fourth_quantil
## premier_quantil 15 2 0 0
## second_quantil 11 21 11 0
## third_quantil 0 3 16 0
## fourth_quantil 0 0 0 22Question 7:
Reload the Boston dataset and standardize the dataset (we did not do this in the Exercise Set, but you should scale the variables to get comparable distances). Calculate the distances between the observations. Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results. (0-4 points)
# Boston is reload
Boston_load <- Boston
# scale of Boston
Boston_scaled <- scale(Boston_load)
# Calculate distance betwwen scaled data points:
distances <- dist(Boston_scaled)
# Visualize the distances using fviz_dist()
library(factoextra)## Welcome! Want to learn more? See two factoextra-related books at https://goo.gl/ve3WBafviz_dist(distances, gradient = list(low = "#00AFBB", mid = "white", high = "#FC4E07"))
# k mean: The k-means algorithm groups observations into clusters based on their similarity, aiming to minimize the variation within clusters while maximizing the variation between clusters. We need to define the number of clusters we need to do the kmean analyse.
# Elbow Method to find optimal number of clusters: Calculate the Within-Cluster-Sum of Squared Errors (WSS) for different values of k, and choose the k for which WSS becomes first starts to diminish. In the plot of WSS-versus-k, this is visible as an elbow.
# The Squared Error for each point is the square of the distance of the point from its representation i.e. its predicted cluster center.
# The WSS score is the sum of these Squared Errors for all the points.
wss <- numeric(10) # Initialize within-cluster sum of squares vector
for (i in 1:10) {
kmeans_model <- kmeans(Boston_scaled, centers = i, nstart = 10)
wss[i] <- kmeans_model$tot.withinss # Store within-cluster sum of squares
}
# Plotting the Elbow Method
plot(1:10, wss, type = "b", xlab = "Number of Clusters", ylab = "Within-cluster Sum of Squares", main = "Elbow Method")
# or we can use the direct function for the Elbow method
kmeans_model_optimal2 <- fviz_nbclust(Boston_scaled, kmeans, method = "wss")
# or we can use the silhuette method
fviz_nbclust(Boston_scaled, kmeans, method = "silhouette")
# Visualize clusters using pairs() or ggpairs()
pairs(Boston_scaled, col = kmeans_model$cluster)
k <- 2 # 2 seems to be the best option according to the Elbow Method and the silhouette method
#Kmean model:
kmeans_model <- kmeans(Boston_scaled, centers = k, nstart = 25)
cluster_assignments <- kmeans_model$cluster
# visualize the clusters thanks to fviz_cluster function
fviz_cluster(kmeans_model, data = Boston_scaled)
library(GGally)
# Combine the scaled data with the cluster assignments (cluster 1 or 2)
clustered_data <- cbind(as.data.frame(Boston_scaled), Cluster = as.factor(cluster_assignments))
# Visualize clusters using ggpairs()
ggpairs(clustered_data, aes(color = Cluster))## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero
## Warning in cor(x, y): the standard deviation is zero## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
# Mean for each cluster and variable:
clustered_data %>%
mutate(Cluster = kmeans_model$cluster) %>%
group_by(Cluster) %>%
summarise_all("mean")## # A tibble: 2 × 15
## Cluster crim zn indus chas nox rm age dis rad
## <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 1 -0.389 0.262 -0.615 0.00291 -0.582 0.245 -0.433 0.454 -0.583
## 2 2 0.724 -0.487 1.14 -0.00541 1.08 -0.455 0.805 -0.844 1.08
## # ℹ 5 more variables: tax <dbl>, ptratio <dbl>, black <dbl>, lstat <dbl>,
## # medv <dbl>